32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 010 001 999 726 227 950 304 746 627 807 615 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 010 001 999 726 227 950 304 746 627 807 615(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 010 001 999 726 227 950 304 746 627 807 615.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 010 001 999 726 227 950 304 746 627 807 615 × 2 = 0 + 0.000 020 003 999 452 455 900 609 493 255 615 23;
  • 2) 0.000 020 003 999 452 455 900 609 493 255 615 23 × 2 = 0 + 0.000 040 007 998 904 911 801 218 986 511 230 46;
  • 3) 0.000 040 007 998 904 911 801 218 986 511 230 46 × 2 = 0 + 0.000 080 015 997 809 823 602 437 973 022 460 92;
  • 4) 0.000 080 015 997 809 823 602 437 973 022 460 92 × 2 = 0 + 0.000 160 031 995 619 647 204 875 946 044 921 84;
  • 5) 0.000 160 031 995 619 647 204 875 946 044 921 84 × 2 = 0 + 0.000 320 063 991 239 294 409 751 892 089 843 68;
  • 6) 0.000 320 063 991 239 294 409 751 892 089 843 68 × 2 = 0 + 0.000 640 127 982 478 588 819 503 784 179 687 36;
  • 7) 0.000 640 127 982 478 588 819 503 784 179 687 36 × 2 = 0 + 0.001 280 255 964 957 177 639 007 568 359 374 72;
  • 8) 0.001 280 255 964 957 177 639 007 568 359 374 72 × 2 = 0 + 0.002 560 511 929 914 355 278 015 136 718 749 44;
  • 9) 0.002 560 511 929 914 355 278 015 136 718 749 44 × 2 = 0 + 0.005 121 023 859 828 710 556 030 273 437 498 88;
  • 10) 0.005 121 023 859 828 710 556 030 273 437 498 88 × 2 = 0 + 0.010 242 047 719 657 421 112 060 546 874 997 76;
  • 11) 0.010 242 047 719 657 421 112 060 546 874 997 76 × 2 = 0 + 0.020 484 095 439 314 842 224 121 093 749 995 52;
  • 12) 0.020 484 095 439 314 842 224 121 093 749 995 52 × 2 = 0 + 0.040 968 190 878 629 684 448 242 187 499 991 04;
  • 13) 0.040 968 190 878 629 684 448 242 187 499 991 04 × 2 = 0 + 0.081 936 381 757 259 368 896 484 374 999 982 08;
  • 14) 0.081 936 381 757 259 368 896 484 374 999 982 08 × 2 = 0 + 0.163 872 763 514 518 737 792 968 749 999 964 16;
  • 15) 0.163 872 763 514 518 737 792 968 749 999 964 16 × 2 = 0 + 0.327 745 527 029 037 475 585 937 499 999 928 32;
  • 16) 0.327 745 527 029 037 475 585 937 499 999 928 32 × 2 = 0 + 0.655 491 054 058 074 951 171 874 999 999 856 64;
  • 17) 0.655 491 054 058 074 951 171 874 999 999 856 64 × 2 = 1 + 0.310 982 108 116 149 902 343 749 999 999 713 28;
  • 18) 0.310 982 108 116 149 902 343 749 999 999 713 28 × 2 = 0 + 0.621 964 216 232 299 804 687 499 999 999 426 56;
  • 19) 0.621 964 216 232 299 804 687 499 999 999 426 56 × 2 = 1 + 0.243 928 432 464 599 609 374 999 999 998 853 12;
  • 20) 0.243 928 432 464 599 609 374 999 999 998 853 12 × 2 = 0 + 0.487 856 864 929 199 218 749 999 999 997 706 24;
  • 21) 0.487 856 864 929 199 218 749 999 999 997 706 24 × 2 = 0 + 0.975 713 729 858 398 437 499 999 999 995 412 48;
  • 22) 0.975 713 729 858 398 437 499 999 999 995 412 48 × 2 = 1 + 0.951 427 459 716 796 874 999 999 999 990 824 96;
  • 23) 0.951 427 459 716 796 874 999 999 999 990 824 96 × 2 = 1 + 0.902 854 919 433 593 749 999 999 999 981 649 92;
  • 24) 0.902 854 919 433 593 749 999 999 999 981 649 92 × 2 = 1 + 0.805 709 838 867 187 499 999 999 999 963 299 84;
  • 25) 0.805 709 838 867 187 499 999 999 999 963 299 84 × 2 = 1 + 0.611 419 677 734 374 999 999 999 999 926 599 68;
  • 26) 0.611 419 677 734 374 999 999 999 999 926 599 68 × 2 = 1 + 0.222 839 355 468 749 999 999 999 999 853 199 36;
  • 27) 0.222 839 355 468 749 999 999 999 999 853 199 36 × 2 = 0 + 0.445 678 710 937 499 999 999 999 999 706 398 72;
  • 28) 0.445 678 710 937 499 999 999 999 999 706 398 72 × 2 = 0 + 0.891 357 421 874 999 999 999 999 999 412 797 44;
  • 29) 0.891 357 421 874 999 999 999 999 999 412 797 44 × 2 = 1 + 0.782 714 843 749 999 999 999 999 998 825 594 88;
  • 30) 0.782 714 843 749 999 999 999 999 998 825 594 88 × 2 = 1 + 0.565 429 687 499 999 999 999 999 997 651 189 76;
  • 31) 0.565 429 687 499 999 999 999 999 997 651 189 76 × 2 = 1 + 0.130 859 374 999 999 999 999 999 995 302 379 52;
  • 32) 0.130 859 374 999 999 999 999 999 995 302 379 52 × 2 = 0 + 0.261 718 749 999 999 999 999 999 990 604 759 04;
  • 33) 0.261 718 749 999 999 999 999 999 990 604 759 04 × 2 = 0 + 0.523 437 499 999 999 999 999 999 981 209 518 08;
  • 34) 0.523 437 499 999 999 999 999 999 981 209 518 08 × 2 = 1 + 0.046 874 999 999 999 999 999 999 962 419 036 16;
  • 35) 0.046 874 999 999 999 999 999 999 962 419 036 16 × 2 = 0 + 0.093 749 999 999 999 999 999 999 924 838 072 32;
  • 36) 0.093 749 999 999 999 999 999 999 924 838 072 32 × 2 = 0 + 0.187 499 999 999 999 999 999 999 849 676 144 64;
  • 37) 0.187 499 999 999 999 999 999 999 849 676 144 64 × 2 = 0 + 0.374 999 999 999 999 999 999 999 699 352 289 28;
  • 38) 0.374 999 999 999 999 999 999 999 699 352 289 28 × 2 = 0 + 0.749 999 999 999 999 999 999 999 398 704 578 56;
  • 39) 0.749 999 999 999 999 999 999 999 398 704 578 56 × 2 = 1 + 0.499 999 999 999 999 999 999 998 797 409 157 12;
  • 40) 0.499 999 999 999 999 999 999 998 797 409 157 12 × 2 = 0 + 0.999 999 999 999 999 999 999 997 594 818 314 24;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 010 001 999 726 227 950 304 746 627 807 615(10) =

0.0000 0000 0000 0000 1010 0111 1100 1110 0100 0010(2)


5. Positive number before normalization:

0.000 010 001 999 726 227 950 304 746 627 807 615(10) =

0.0000 0000 0000 0000 1010 0111 1100 1110 0100 0010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 010 001 999 726 227 950 304 746 627 807 615(10) =

0.0000 0000 0000 0000 1010 0111 1100 1110 0100 0010(2) =

0.0000 0000 0000 0000 1010 0111 1100 1110 0100 0010(2) × 20 =

1.0100 1111 1001 1100 1000 010(2) × 2-17


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -17

Mantissa (not normalized):
1.0100 1111 1001 1100 1000 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-17 + 2(8-1) - 1 =

(-17 + 127)(10) =

110(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 110 ÷ 2 = 55 + 0;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

110(10) =

0110 1110(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 010 0111 1100 1110 0100 0010 =

010 0111 1100 1110 0100 0010


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 1110

Mantissa (23 bits) =
010 0111 1100 1110 0100 0010


The base ten decimal number 0.000 010 001 999 726 227 950 304 746 627 807 615 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0110 1110 - 010 0111 1100 1110 0100 0010

More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation:

» Number 0.000 010 001 999 726 227 950 304 746 627 807 614 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

» Number 0.000 010 001 999 726 227 950 304 746 627 807 616 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111